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Steel block under triaxial loads: dimension and volume change

Short answer

Compute stresses σ_x=15.56, σ_y=-30.00, σ_z=33.33 MPa. Use generalized Hooke’s law with E=200 GPa, μ=0.30 to find strains, then dimension changes: +0.0073 mm in x, -0.0335 mm in y, +0.0565 mm in z. Volume change = +170.0 mm³.

A block of steel 300 mm×150 mm×100 mm300\ \text{mm}\times150\ \text{mm}\times100\ \text{mm} is subjected to axial loads as shown in figure below. Find the change in the dimensions of the bar and change in volume for the material of the block. Take Es=200 GN/m2E_s = 200\ \text{GN/m}^2 and poisson's ratio (σ\sigma) = 0.30. [Figure: block with dimensions 300 mm (along Z), 150 mm (along Y) and 100 mm (along X); forces: 500 kN along Z (pulling at the top and bottom), 700 kN along X (tensile at both ends), 900 kN along Y (compressive).]

Take x = 100 mm, y = 150 mm, z = 300 mm. Tension is positive. E=200×103E = 200\times10^3 N/mm², μ=0.30\mu = 0.30.

Stresses (load ÷ area of the loaded face)

σz=500×103100×150=33.333 N/mm2 (tension)σx=700×103150×300=15.556 N/mm2 (tension)σy=−900×103100×300=−30.000 N/mm2 (compression)\begin{aligned} \sigma_z &= \frac{500\times10^3}{100\times150} = 33.333\ \text{N/mm}^2\ (\text{tension})\\ \sigma_x &= \frac{700\times10^3}{150\times300} = 15.556\ \text{N/mm}^2\ (\text{tension})\\ \sigma_y &= -\frac{900\times10^3}{100\times300} = -30.000\ \text{N/mm}^2\ (\text{compression}) \end{aligned}

Strains (generalised Hooke's law)

εx=1E[σx−μ(σy+σz)]=1200×103[15.556−0.3(−30.000+33.333)]=7.2778×10−5εy=1E[σy−μ(σx+σz)]=−2.2333×10−4εz=1E[σz−μ(σx+σy)]=1.8833×10−4\begin{aligned} \varepsilon_x &= \frac{1}{E}\left[\sigma_x - \mu(\sigma_y+\sigma_z)\right] = \frac{1}{200\times10^3}\left[15.556 - 0.3(-30.000+33.333)\right] = 7.2778\times10^{-5}\\ \varepsilon_y &= \frac{1}{E}\left[\sigma_y - \mu(\sigma_x+\sigma_z)\right] = -2.2333\times10^{-4}\\ \varepsilon_z &= \frac{1}{E}\left[\sigma_z - \mu(\sigma_x+\sigma_y)\right] = 1.8833\times10^{-4} \end{aligned}

Change in dimensions

δx=εx (100)=+0.00728 mmδy=εy (150)=−0.03350 mmδz=εz (300)=+0.05650 mm\begin{aligned} \delta_x &= \varepsilon_x\,(100) = +0.00728\ \text{mm}\\ \delta_y &= \varepsilon_y\,(150) = -0.03350\ \text{mm}\\ \delta_z &= \varepsilon_z\,(300) = +0.05650\ \text{mm} \end{aligned}

Change in volume

εv=εx+εy+εz=3.7778×10−5,δV=εvV=3.7778×10−5×(100×150×300)=+170.00 mm3\varepsilon_v = \varepsilon_x+\varepsilon_y+\varepsilon_z = 3.7778\times10^{-5},\qquad \delta V = \varepsilon_v V = 3.7778\times10^{-5}\times(100\times150\times300) = +170.00\ \text{mm}^3

Check with εv=(1−2μ)E(σx+σy+σz)=0.4200×103(18.889)=3.7778×10−5\varepsilon_v = \dfrac{(1-2\mu)}{E}(\sigma_x+\sigma_y+\sigma_z) = \dfrac{0.4}{200\times10^3}(18.889) = 3.7778\times10^{-5}.

Answer: δx=+0.0073\delta_x = +0.0073 mm, δy=−0.0335\delta_y = -0.0335 mm, δz=+0.0565\delta_z = +0.0565 mm; δV=+170.0\delta V = +170.0 mm³ (an increase).

✓ Answer checked by Claude Opus 5.5 (high effort)

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